# Deriving Biot-Savart's Law, kinda a math problem not really physics

But your theta is wrong, at least in my picture it is wrong.

jhae2.718
Gold Member
x is the green line, right? Then if theta is between r and y (the line ds is along), then x = r*sin(theta)

yeah, you had $$<sin\theta, cos\theta,0>$$

jhae2.718
Gold Member
Your vector is in <x,y,z> form, right? Then x = rsin(theta), y = rcos(theta), and z = 0.

Wait you may be onto something...*goes and fix tex*

Okay! Tex is fixed on first page, made some other minor substition fixed. BUt my x^2 still remains..

jhae2.718
Gold Member
Use the approach you did in post #3...

I also TeX-ed up my solution; I'll post it in a little bit...

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SammyS
Staff Emeritus
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Okay! Tex is fixed on first page, made some other minor substitution fixed. But my x^2 still remains..
The last I looked, you did have inconsistencies between the way θ is drawn in your figure and the way it's used in some of your equations.

$$d\vec{s} = <0,d\vec{y},0>$$  should be  $$d\vec{s} = <0,dy,0>$$

$$\hat{r} = <\cos\theta,\sin\theta,0>$$  should be  $$\hat{r} = <\sin\theta,\cos\theta,0>$$

In general  $$\left|\vec{A}\times\vec{C}\right|=\left|\vec{A}\right|\left|\vec{C}\right|\sin\theta$$ so your cross product should have sin θ, not cos θ.

Jumping down to one of your equations for B:

$$\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-cos\theta d\vec{y} \hat{k}}{x^2 + y^2}$$

After the above corrections and sin θ = x/r, you should have:

$$\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-\sin\theta dy \hat{k}}{x^2 + y^2}$$
$$=-\,\frac{x\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x^2 + y^2)^{3/2}}$$

x comes out of the integral because you're integrating over y, not x.​

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This is going to sound unrelated, but how did you make that pdf that is so compatiable with LaTeX?

SammyS
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Here's my solution:
View attachment 33843
I didn't do the integration, but my solution agrees with this.

But, I do think there is a problem with both solutions.

It's late, so I'll look at it tomorrow.

The concern I have with my solution, is that the variable, y, was used in two distinct ways. The primary way was as the variable of integration. The other way y was used was as a coordinate for the point, P.

I intend to redo my solution using (x0, y0) for the coordinates of P.

.

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jhae2.718
Gold Member
This is going to sound unrelated, but how did you make that pdf that is so compatiable with LaTeX?

I used LaTeX :)

MikTeX 2.9 on Windows.

Source:
Code:
\documentclass[10pt,letterpaper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{fullpage}
\usepackage{graphicx}
\usepackage{float}
\usepackage{xfrac}
\newcommand{\custpic}[4]{\begin{figure}[H]
\begin{center}
\includegraphics[width=#2\textwidth]{#1}
\vspace{-15pt}
\caption{#3}
\vspace{-20pt}
\label{#4}
\end{center}
\end{figure}}

\begin{document}
\custpic{pf5}{.75}{Problem diagram}{pf}

$$\label{bs} d\vec{B} = \frac{\mu_0}{4\pi}\frac{id\vec{s} \times \hat{r}}{r^2}$$
From the diagram:
\begin{align*}
r = \sqrt{x^2 + y^2} \\
x = r\sin(\theta) \\
y = r\cos(\theta) \\
d\vec{s} = d\vec{y}
\end{align*}
Substitute into Eq. (\ref{bs}):
$$d\vec{B} = \frac{\mu_0}{4\pi}\frac{id\vec{y} \times \hat{r}}{x^2 + y^2}$$
Taking the cross product, $id\vec{y} \times \hat{r} = dy\sin(\theta) (-\hat{k})$. Substitute in:
$$d\vec{B} = \frac{\mu_0 i}{4\pi}\frac{dy\sin(\theta)}{x^2 + y^2} (-\hat{k})$$
From above, $\sin(\theta) = \dfrac{x}{r}$:
$$d\vec{B} = \frac{\mu_0 i}{4\pi}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})$$
Find $\vec{B}$ by taking the integral over the infinite line:
$$\vec{B} = \frac{\mu_0 i}{4\pi}\int\limits_{-\infty}^{\infty}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})$$
$$\vec{B} = 2 \frac{\mu_0 i}{4\pi}\int\limits_{0}^{\infty}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})$$
Integrate (I used Wolfram$|$Alpha) to get:
$$\vec{B} = 2 \frac{\mu_0 i}{4\pi}\left[\frac{y}{x\sqrt{x^2 + y^2}}\right]_{0}^{\infty} (-\hat{k})$$
The indefinite integral evaluates to:
$$\vec{B} = \frac{\mu_0 i}{2\pi x} (-\hat{k})$$
We can confirm this using Ampere's Law (left as an exercise to the reader...)
\end{document}

It's getting late here, I will recheck my math tomorrow, thank you to both of you very much. That sentence did not make any sense, but hopefully you understood my appreciation.

How did you save the file so perfectly? Whenever I saved it as a pdf, and I try reopen it it says it can't read it.

SammyS
Staff Emeritus
Homework Helper
Gold Member
...
The concern I have with my solution, is that the variable, y, was used in two distinct ways. The primary way was as the variable of integration. The other way y was used was as a coordinate for the point, P.
...
Simply use x0 for x, and use (y0 - y) for y.

$$\vec{B}(x_0,\,y_0)=-\,\frac{x_0\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x_0^2 + (y_0-y)^2)^{3/2}}$$

I think you may find it handy to evaluate the integral from y0 to y .

jhae2.718
Gold Member
How did you save the file so perfectly? Whenever I saved it as a pdf, and I try reopen it it says it can't read it.

You need to run latex on it.

Simply use x0 for x, and use (y0 - y) for y.

$$\vec{B}(x_0,\,y_0)=-\,\frac{x_0\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x_0^2 + (y_0-y)^2)^{3/2}}$$

I think you may find it handy to evaluate the integral from y0 to y .

That is an excellent point, and is something I just glossed over.

WHat dose that mean "run latex" over it...?

jhae2.718
Gold Member
If you have a version of LaTeX installed, open up the command prompt, change directory to the location where you have saved the file, and type:
Code:
pdflatex filename.tex
replacing filename with the actual filename.

I am guessing LaTeX is not the same thing as MikTeX lol

jhae2.718
Gold Member
MikTeX is a LaTeX distribution. http://docs.miktex.org/faq/faq.html

Try a LaTeX editor like TeXniCenter, Texmaker, etc to actually write/compile the code if you don't like the CLI.